Capacitor behavior with constant voltage and decreasing separation

Steps:

• Battery maintains constant potential difference V across plates.
• Capacitance C = ε₀A/d increases as separation d decreases.
• Charge Q = CV increases due to higher C and fixed V.
• Electric field E = V/d increases as d decreases with V constant.

Why B is correct:

• Battery fixes V constant; E = V/d shows E increases with smaller d (from field definition for parallel plates).

Why the others are wrong:

• A: E increases, not decreases, per E = V/d.
• C: E increases, not decreases; V constant, not increasing.
• D: V constant from battery, not increasing.

Final answer: B